# Li-Ion battery charged by anemometer

Hi guys,

I am trying to charge this Li-Ion battery (https://www.sparkfun.com/products/339) with a small dc generator.

It is attached to an anemometer and i can get 100 mV and 20 mA in a normal windy day.

I have this dc/dc converter to boost the 100mV to 4.1V so i can charge the battery (http://www.linear.com/product/LTC3108)

What i need now is to manage the battery so it won't overcharge and don't let it discharge completely.

This circuit is to power an Arduino Fio, so i need low power.

Can someone help me?

You want to check out Li-Ion Charger, since they need speacial care when it comes to charging. Lead batteries are less critical...

Regards b

From the product page of your Sparkfun battery:

"Battery includes built-in protection against over voltage, over current, and minimum voltage."

Sparkfun also provides the datasheet for that protection circuit.

Some basic maths needed here. The anemometer produces 100 mV @ 20 mA which is a total power output of 2 milliwatts. You are trying to charge a 1000mah 3.7 v battery with a 2 milliwatt charger. Its going to take well over 2000 hours to charge the battery.

I can get 150 mV and 150 mA with 1 impulse... The arduino Fio has very little consumes, so i don't really need to charge the battery completely, i only need to collect some energy through a long period of time to keep the it working in sleep mode and some peak points to get information.

Imagine i would get 150 mV and 150 mA in a constant source. It would give me 22.5 mW of energy. To boost that tension to 4.1 V, what current would i get out of the step up converter?

How are you measuring the output power of the generator. You indicate that you can get 150 mV @ 150 mA from the generator. Did you measure this into some kind of resistive load?